Question: Simplify; express your answer in exponential form. Assume $r\neq 0, y\neq 0$. $\dfrac{{(r)^{-1}}}{{(r^{2}y^{-1})^{-4}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${r}$ to the exponent ${-1}$ . Now ${1 \times -1 = -1}$ , so ${(r)^{-1} = r^{-1}}$ In the denominator, we can use the distributive property of exponents. ${(r^{2}y^{-1})^{-4} = (r^{2})^{-4}(y^{-1})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r)^{-1}}}{{(r^{2}y^{-1})^{-4}}} = \dfrac{{r^{-1}}}{{r^{-8}y^{4}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-1}}}{{r^{-8}y^{4}}} = \dfrac{{r^{-1}}}{{r^{-8}}} \cdot \dfrac{{1}}{{y^{4}}} = r^{{-1} - {(-8)}} \cdot y^{- {4}} = r^{7}y^{-4}$.